Electron configurations follow the order of sublevels on the periodic table.

Transcription

1 Electron configurations follow the order of sublevels on the periodic table. 1

2 The periodic table consists of sublevel blocks arranged in order of increasing energy. Groups 1A(1)-2A(2) = s level Groups 3A(13)-8A(18) = p level Groups 3B(3) to 2B(12) = d level Lanthanides/Actinides = f level 2

3 To write an electron configuration using Sublevel blocks, locate the element on the periodic table starting with H in 1s,write each sublevel block in order going from left to right across each period write the number of electrons in each block 3

4 Solution Period 1 1s block 1s 2 Period 2 2s 2p blocks 2s 2 2p 6 Period 3 3s 3p blocks 3s 2 3p 2 (Si) Writing all the sublevel blocks in order gives 1s 2 2s 2 2p 6 3s 2 3p 2 4

5 A. 1s 2 2p 5 25% 25% 25% 25% B. 1s 2 2s 2 2p 6 C. 1s 2 2s 2 2p 3 D. 1s 2 2s 2 2p 4 1s22p5 1s22s22p6 1s22s22p3 1s22s22p4 5

6 A. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 25% 25% 25% 25% B. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 C. 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 D. 1s 2 2s 2 2p 6 3s 2 3p 5 4s 2 3d 1 1s22s22p63s23p64s2 1s22s22p63s23p64s1 1s22s22p63s23p63d2 1s22s22p63s23p54s23d1 6

7 For 3d transition metals, 3d has lower energy after the electrons is filled. 3d and 4s are close in energy 1s 2s 2p 3s 3p 4s 3d Ar 1s 2 2s 2 2p 6 3s 2 3p 6 K 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 Ca 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 Sc 1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2 Ti 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 4s 2 7

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10 A. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 25% 25% 25% 25% B. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 5 C. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 3 D. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2 1s22s2 2p63s2 3p6 4s2 1s22s2 2p63s2 3p6 4s2 3d5 1s22s2 2p63s2 3p6 4s2 3d3 1s22s2 2p63s2 3p6 4s2 3d2 10

11 Solution Period 1 1s block 1s 2 Period 2 2s 2p blocks 2s 2 2p 6 Period 3 3s 3p blocks 3s 2 3p 6 Period 4 4s 3d blocks 4s 2 3d 2 (at Ti) Writing all the sublevel blocks in order gives 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2 11

12 A. Na B. S C. K D. Ca 0% 0% 0% 0% Na S K Ca 12

13 100% A. 3p 6 4s 2 B. 4s 2 4d 7 C. 4s 2 3d 7 0% 0% 3p64s2 4s24d7 4s23d7 13

14 The last filled principal energy level is called the valence level, or valence shell which is the outermost sublevels that are highest in energy. The valence electrons occupy orbitals in the valence level All the other electrons are called core electrons, or inner electrons. determine the chemical properties of an element are related to the group number of the element Example: Phosphorus has 5 valence electrons 5 valence electrons P Group 5A(15) 1s 2 2s 2 2p 6 3s 2 3p 3 14

15 All the elements in a group have the same number of valence electrons. Example: Be 1s 2 2s 2 Mg 1s 2 2s 2 2p 6 3s 2 Ca 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 Sr 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 Can you see a pattern? Elements in Group 2A (2) have two (2) valence electrons. 15

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17 State the number of valence electrons for each of following: A. oxygen 1) 4 2) 6 3) 8 B. aluminum 1) 13 2) 3 3) 1 C. chlorine 1) 2 2) 5 3) 7 Solution: A. 2) 6 B. 2) 3 C. 3) 7 17

18 State the number of valence electrons for each. A. 1s 2 2s 2 2p 6 3s 2 3p 3 B. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4 C. 1s 2 2s 2 2p 5 Solution: A. 5 B. 6 C. 7 18

19 Noble gas-core abbreviated electron configurations are often used for elements with many electrons. Notice that iron s electron configuration starts out with argon s electron configuration, but ends differently: Fe 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 Ar 1s 2 2s 2 2p 6 3s 2 3p 6 We use the symbol [Ar] to represent argon s electron configuration: Fe [Ar] 4s 2 3d 6 argon core + valence electrons 7-19

20 In atoms, the number of electrons is equal to the number of protons, which is the atomic number. In ions, the number of electrons does not equal the atomic number. We must add or subtract electrons, depending on whether the ion is an anion or cation. 7-20

21 An example problem: Write the electron configuration for Na + : Na + has a positive charge of 1; therefore, we need to subtract 1 electron from the total number of electrons, 11. Na + has 10 electrons and is isoelectronic with Ne. 1s 2 2s 2 2p

22 Write the electron configuration in spdf notation and noble gas-core abbreviated electronic configuration for the following ions. 1. Br - 2. N 3-3. K + 4. Sr S 2-6. Ni

23 Write the electron configuration in spdf noation and noble gas core abbreviated for the following ions. 1. Br - 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 [Kr] 2. N 3-1s 2 2s 2 2p 6 [Ne] 3. K + 1s 2 2s 2 2p 6 3s 2 3p 6 [Ar] 4. Sr 2+ 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 [Kr] 5. S 2-1s 2 2s 2 2p 6 3s 2 3p 6 [Ar] 6. Ni 2+ 1s 2 2s 2 2p 6 3s 2 3p 6 3d 8 [Ar] 3d

24 Electron configurations are related to the following properties: Relative reactivity Atomic radii (atomic size) Ionization Energy (tendency to lose electrons) 7-24

25 From the activity series we see that the most active metals are those in Groups IA and IIA. The more active the metal, the more easily it loses electrons to form ions. Figure

26 The alkali metals form +1 ions since they all have 1 valence electron. Oxides of the alkali metals: Li 2 O, Na 2 O, K 2 O, Rb 2 O, Cs 2 O The alkaline earth metals form +2 ions since they all have 2 valence electrons. Oxides of the alkaline earth metals: BeO, MgO, CaO, SrO, BaO 7-26

27 Ionization energy is the energy it takes to remove a electron from an atom or ion in the gas state. 27

28 In general, for Main group elements: the ionization energy increase in going left to right across a in period. Except, IE(group 3A) < IE(group 2A) and IE(group 6A) < IE(group 5A) the ionization energy decreases in going down a group. 28

29 Select the element in each pair with the higher ionization energy. A. Li or K B. K or Br C. P or Cl Solution A. Li B. Br C. Cl 29

30 What do these values tell us about the stability of core (inner) electrons? 7-30

31 Atomic size is usually reported atomic radius covalent radius: half the distance between identical nuclei of the neighboring atoms of the same element in a molecules Metallic radius: is based on the distance of closest approach of adjacent atoms in a solid metal. 31

32 The atomic radius in general increases going down each group as the number of energy levels increases Except : Al > Ga 32

33 The atomic radius in general decreases going from left to right across a period because electrons are held closer to the nucleus by the increasingly greater charge of the nucleus. 33

34 Select the element in each pair with the larger atomic radius. A. Li or K B. K or Br C. P or Cl 34

35 Select the element in each pair with the larger atomic radius. A. K is larger than Li B. K is larger than Br C. P is larger than Cl 35

36 A positive ion has lost its valence electrons is smaller than the corresponding neutral atom Example: Group 1A and metals ions about half the size of its corresponding neutral atoms 36

37 The sodium ion Na + forms when the Na atom loses one electron from the third energy level is smaller than a Na atom 37

38 A negative ion increases the number of valence electrons is larger than the corresponding neutral atom Example: Group 7A and nonmetal ions about twice the size of its corresponding neutral atoms 38

39 The fluoride ion F - forms when a valence electron is added has increased repulsions due to the added valence electron is larger than a F atom 39

40 For any isoelectronic (same number of electrona) series, as the number of protons increases, the ion size decreases. Figure from p

41 The general trend for ionic size (or radius) is for ionic size to increase from top to bottom in the periodic table. For cations, as the charge increases, the ionic size decreases. For anions, as the charge increases (becomes more negative), the ionic size increases. 7-41

42 1. Which is larger in each of the following? A. K or K + B. Al or Al 3+ C. S 2- or S 2. Which is smaller in each of the following? A. N 3- or N B. Cl or Cl - C. Sr 2+ or Sr 42

43 1. Which is larger in each of the following? A. K > K + B. Al > Al 3+ C. S 2- > S 2. Which is smaller in each of the following? A. N < N 3- B. Cl < Cl - C. Sr 2+ < Sr 43

44 Ability of an atom to attract bonding electrons Proposed by Linus Pauling in the early 1930 s A difference in electronegativity between the atoms in a covalent bond results in: A polar covalent bond Increased ionic character The greater the difference in electronegativity, the greater the ionic character and the more polar the bond that joins the atoms. Decreased bond length and increased bond strength No difference in electronegativity between atoms in a covalent bond results in a nonpolar covalent bond. 8-44

45 Electronegativity Values Figure

46 The difference in electronegativity between metals and nonmetals is so large, that the electrons are transferred, not shared. The greater the electronegativity difference, the more polar the bond. Si-F > N-F> O-F >F-F What partial charges go on each atom? Figure